Question

Choose the correct option from the following

• A. KF is more covalent than KI
• B. \(\mathrm{SnCl_4}\) is less covalent than \(\mathrm{SnCl_2}\)
• C. LiF is more covalent than KF
• D. \(\mathrm{ZnCl_2}\) is less covalent than NaCl

Hint:

According to Fajan's rule, smaller cations and larger anions increase covalent character.

Answer 1

The Correct Option is C

Step 1: Recall Fajan's rule.
According to Fajan's rule, covalent character increases when the cation is small and has high polarising power.
Also, covalent character increases when the anion is large and easily polarisable.

Step 2: Compare LiF and KF.
Both LiF and KF contain the same anion, \(\mathrm{F^-}\).
So, the covalent character depends mainly on the polarising power of the cation.
The size of \(\mathrm{Li^+}\) is smaller than the size of \(\mathrm{K^+}\).
Therefore, \(\mathrm{Li^+}\) has greater polarising power than \(\mathrm{K^+}\).
Hence, LiF has more covalent character than KF.

Step 3: Check the other options.
KI is more covalent than KF because \(\mathrm{I^-}\) is larger and more polarisable than \(\mathrm{F^-}\).
So, option (1) is incorrect.
\(\mathrm{SnCl_4}\) is more covalent than \(\mathrm{SnCl_2}\) because \(\mathrm{Sn^{4+}}\) has greater polarising power than \(\mathrm{Sn^{2+}}\).
So, option (2) is incorrect.
\(\mathrm{ZnCl_2}\) is more covalent than NaCl because \(\mathrm{Zn^{2+}}\) has higher charge and greater polarising power than \(\mathrm{Na^+}\).
So, option (4) is incorrect.

Step 4: Final conclusion.
Hence, the correct statement is
\[ \boxed{\text{LiF is more covalent than KF}} \]